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Thread: [SOLVED] Sine Question

  1. #1
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    [SOLVED] Sine Question

    On a sunny day, a tower casts a shadow 35.2m long. At the same time, a 1.3m parking meter that is nearby casts a shadow 1.8m long. How high is the tower to the nearest tenth of a metre?

    Triangle formed by parking meter:
    $\displaystyle c^2 = a^2 + b^2$
    $\displaystyle c^2 = 1.3^2 + 1.8^2$
    $\displaystyle c^2 = \frac{493}{100}$
    $\displaystyle c = \frac{\sqrt{493}}{10}$

    $\displaystyle SinA = \frac{O}{H}$
    $\displaystyle SinA = \frac{1.3}{\frac{\sqrt{493}}{10}}$
    $\displaystyle Sin^{-1}\frac{1.3}{\frac{\sqrt{493}}{10}} = A$
    $\displaystyle 35.84^{\circ} = A$

    Triangle formed by tower (Total triangle):
    $\displaystyle \frac{a}{SinA} = \frac{b}{SinB}$
    $\displaystyle \frac{a}{Sin35.84} = \frac{37}{Sin54.16}$
    $\displaystyle a = Sin35.84(\frac{37}{Sin54.16})$
    $\displaystyle a = 26.7m$

    *The $\displaystyle 37m$ value was obtained by adding $\displaystyle 35.2m$ and $\displaystyle 1.8m$.
    *The $\displaystyle 54.16^{\circ}$ value was obtained by subtracting $\displaystyle 90deg$ and $\displaystyle 35.84deg$ from $\displaystyle 180deg$.

    Therefore, the height of the tower is $\displaystyle 26.7m$.

    Textbook answer: $\displaystyle 25.4m$.

    Where did I make the mistake?
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  2. #2
    A riddle wrapped in an enigma
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    Hi RogueDemon,

    Actually, you made this way harder than it is.

    You have two similar right triangles.

    Set up a proportion using the ratios of the heights and shadows.

    $\displaystyle \frac{bldg \:\: ht.}{1.3}=\frac{35.2}{1.8}$
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  3. #3
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    Thanks very much!
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